$A$ is a $n \times n$ real matrix.
$A_{ij} = \begin{cases} \phantom{-}b_{i}c_{j} & \text{if } i = j \\
-b_{i}c_{j} & \text{if } i \ne j \end{cases}$How to simplify $\det(A)$?
Update:
Can I simplify the determinant with elementary row and column operations as described at http://www.maths.nuigalway.ie/~rquinlan/MA203/section2-5.pdf?
- Divide each row by $b_{i}$ (elementary row operation)
- Divide each column by $c_{j}$ (elementary column operation)
$\det(A) = \left(\prod_{i=1}^{n} b_{i} \right) \left(\prod_{j=1}^{n} c_{j} \right) \det(S)$
where $S_{ij} = \begin{cases} + 1 & \text{if } i = j \\
-1 & \text{if } i \ne j \end{cases}$
So, the problem reduces to finding $\det(S)$.
Best Answer
For $n\ge 3$ there is a clear pattern. For $n=3$ we have $$ \det(A)=- 4b_1b_2b_3c_1c_2c_3. $$ For $n=4$ we have $$ \det(A)=- 16b_1b_2b_3b_4c_1c_2c_3c_4. $$ So we should have $\det(A)=-f(n)\prod_{i=1}^n b_i\prod_{i=1}^ n c_i$ for all $n\ge 3$ with a positive integer $f(n)$. We have $$ f(3)=4,\; f(4)=16,\; f(5)=48,\; f(6)=128,\; f(7)=320. $$ Conjecture: $f(n)=2^n(n-1)$.